x86 - lea assembly instruction

Question

I Just want to make sure I am reading this right:

movl 12(%ebp), %edx
leal (%edx, %edx, 4), %eax

I read the first line as: edx = [epb + 12], and the second line as: eax = edx + edx*4

Can anybody clarify?

Also, what if I had the following two lines:

leal (%edx, %edx, 4), %eax
leal (%edx, %edx, 2), %eax

Once the second line is executed, would the eax register be overwritten?

And the eax = edx + edx*4 is multiplying the address by 4? Or the contents of the address by 4?

Answer 1

The instruction movl 12(%ebp), %edx means: edx = [ebp + 12]. This is a memory reference (a read operation) to the address ebp + 12 whose contents (a double word) are read to edx register.

The instruction leal (%edx, %edx, 4), %eax means: eax = edx * 5 (which is a simplification of eax = edx + edx * 4). The leal instruction doesn't do memory references. It only performs arithmetic with registers.

As an answer to your second question: Yes, eax would be overwritten because the instruction leal (%edx, %edx, 2), %eax means eax = edx * 3 which is different from the first instruction, eax = edx * 5.